I am trying to learn some regular expressions in Python. The following does not produce the output I expected:
with open('ex06-11.html') as f:
a = re.findall("<div[^>]*id\\s*=\\s*([\"\'])header\\1[^>]*>(.*?)</div>", f.read())
# output: [('"', 'Some random text')]
The output I was expecting (same code, but without the backreference):
with open('ex06-11.html') as f:
print re.findall("<div[^>]*id\\s*=\\s*[\"\']header[\"\'][^>]*>(.*?)</div>", f.read())
# output: ['Some random text']
The question really boils down to: why is there a quotation mark in my first output, but not in my second? I thought that ([abc]) ... //1 == [abc] ... [abc]. Am I incorrect?
From the docs on re.findall:
If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.
If you want the entire match to be returned, remove the capturing groups or change them to non-capturing groups by adding ?: after the opening paren. For example you would change (foo) in your regex to (?:foo).
Of course in this case you need the capturing group for the backreference, so your best bet is to keep your current regex and then use a list comprehension with re.finditer() to get a list of only the second group:
regex = re.compile(r"""<div[^>]*id\s*=\s*(["'])header\1[^>]*>(.*?)</div>""")
with open('ex06-11.html') as f:
a = [m.group(2) for m in regex.finditer(f.read())
A couple of side notes, you should really consider using an HTML parser like BeautifulSoup instead of regex. You should also use triple-quoted strings if you need to include single or double quotes within you string, and use raw string literals when writing regular expressions so that you don't need to escape the backslashes.
The behaviour is clearly documented. See re.findall:
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found.
If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.
So, if you have a capture group in your regex pattern, then findall method returns a list of tuple, containing all the captured groups for a particular match, plus the group(0).
So, either you use a non-capturing group - (?:[\"\']), or don't use any group at all, as in your 2nd case.
P.S: Use raw string literals for your regex pattern, to avoid escaping your backslashes. Also, compile your regex outside the loop, so that is is not re-compiled on every iteration. Use re.compile for that.
When I asked this question I was just starting with regular expressions. I have since read the docs completely, and I just wanted to share what I found out.
Firstly, what Rohit and F.J suggested, use raw strings (to make the regex more readable and less error-prone) and compile your regex beforehand using re.compile. To match an HTML string whose id is 'header':
s = "<div id='header'>Some random text</div>"
We would need a regex like:
p = re.compile(r'<div[^>]*id\s*=\s*([\"\'])header\1[^>]*>(.*?)</div>')
In the Python implementation of regex, a capturing group is made by enclosing part of your regex in parentheses (...). Capturing groups capture the span of text that they match. They are also needed for backreferencing. So in my regex above, I have two capturing groups: ([\"\']) and (.*?). The first one is needed to make the backreference \1 possible. The use of a backreferences (and the fact that they reference back to a capturing group) has consequences, however. As pointed out in the other answers to this question, when using findall on my pattern p, findall will return matches from all groups and put them in a list of tuples:
print p.findall(s)
# [("'", 'Some random text')]
Since we only want the plain text from between the HTML tags, this is not the output we're looking for.
(Arguably, we could use:
print p.findall(s)[0][1]
# Some random text
But this may be a bit contrived.)
So in order to return only the text from between the HTML tags (captured by the second group), we use the group() method on p.search():
print p.search(s).group(2)
# Some random text
I'm fully aware that all but the most simple HTML should not be handled by regex, and that instead you should use a parser. But this was just a tutorial example for me to grasp the basics of regex in Python.
Related
I am learning about regular expressions. I need to match things in a parenthesis group followed by some pattern that I define. When I try this with regular expressions (in Python), it only returns the part in parentheses that it matched, but not the pattern which follows it. An example should clarify:
import re
s = "texttoignore_ABCABC12345_moretexttoignore"
re.findall("(ABC)+\d+", s)
When I speak of the parenthesis group, in the example above this is the "(ABC)+" part. What I intend is for it to look for one or more repetitions of the pattern in parentheses (in this case "ABC"), then the pattern after.
The problem is this: it does not return the pattern after. (In this example, it would return 'ABC', but I would want 'ABCABC12345' or 'ABC12345' or better yet '12345')
How can you include the part after the parentheses in the return value? Is this something about regular expressions or is it specific to this Python method?
Thanks!
John
The "problem" here is that rather specific behavior of re.findall
If one or more groups are present in the pattern, return a
list of groups; this will be a list of tuples if the pattern
has more than one group.
There are a few options you have here. Either make your group non-capturing:
>>> re.findall("(?:ABC)+\d+", s)
['ABCABC12345']
or use re.finditer:
>>> [m.group(0) for m in re.finditer("(ABC)+\d+", s)]
['ABCABC12345']
If you only want to find the pattern once, then #Jkdc's approach from the comments works fine.
>>> re.search("(ABC)+\d+", s).group()
'ABCABC12345'
I'm trying to create a regex in Python to match everything not in mustache brackets.
For example, this string:
This is an {{example}} for {{testing}}.
should produce this list of strings:
["This is an ", " for ", "."]
when used with re.findall.
For my mustache-matching regex, I am using this: {{(.*?)}}.
It seems like it should just be a simple matter of negating the above pattern, but I can't get it to work properly. I'm testing using: http://pythex.org
Thanks.
You need split
re.split('{{.*?}}', s)
#Andrey solution is the most simple and clearly the way to go. You have another possible way with findall because when the pattern contains a capture group, it returns only the capture group content:
re.findall(r'(?:{{.*?}})*([^{]+)', s)
So, if you start the pattern with an optional non-capturing group for curly brackets parts followed by a capture group for other content, findall returns only the capture group content and all curly brackets parts are consumed.
This question already has answers here:
re.findall behaves weird
(3 answers)
Closed 3 years ago.
I am trying to match two string variables, and would like to catch multiple matches. re.findall seems like the obvious choice for this task, but it doesn't appear to be working the way I would expect it to. The following is an example:
a = 'a(pp)?le'
b = 'ale, apple, apol'
match = re.findall(a,b)
match
['','pp']
However, when I apply the same variables to re.search, it recognizes the embedded regular expression within the string, and picks up the first match:
match = re.search(a,b)
match.group()
'ale'
Can anyone explain why re.findall is not working in this instance? I would expect the following:
match = re.findall(a,b)
match
['ale','apple']
Thanks!
You are using a capturing group, wheras you want a non-capturing group:
a = 'a(?:pp)?le'
As stated in the docs (...) in a regex will create a "capturing group" and the result of re.findall will be only what is inside the parens.
If you just want to group things (e.g. for the purpose of applying a ?) use (?:...)which creates a non-capturing group. The result of re.findall in this case will be the whole regex (or the largest capturing group).
The key part of the re.findall docs are:
If one or more groups are present in the pattern, return a list of groups
this explains the difference in results between re.findall and re.search.
Let me quote the Python docs about re.findall():
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.
And this is what your expression a(pp)?le does. It matches the content in your group, i.e. pp. You can always disable this special behavior of a group by taking a non-capturing group (?:...).
I've currently written a nooby regex pattern which involves excessive use of the "(" and ")" characters, but I'm using them for 'or' operators, such as (A|B|C) meaning A or B or C.
I need to find every match of the pattern in a string.
Trying to use the re.findall(pattern, text) method is no good, since it interprets the parenthesis characters as indexing signifiers (or whatever the correct jargon be), and so each element of the produced List is not a string showing the matched text sections, but instead is a tuple (which contain very ugly snippets of pattern match).
Is there an argument I can pass to findall to ignore paranthesis as indexing?
Or will I have to use a very ugly combination of re.search, and re.sub
(This is the only solution I can think of; Find the index of the re.search, add the matched section of text to the List then remove it from the original string {by using ugly index tricks}, continuing this until there's no more matches. Obviously, this is horrible and undesirable).
Thanks!
Yes, add ?: to a group to make it non-capturing.
import re
print re.findall('(.(foo))', "Xfoo") # [('Xfoo', 'foo')]
print re.findall('(.(?:foo))', "Xfoo") # ['Xfoo']
See re syntax for more information.
re.findall(r"(?:A|B|C)D", "BDE")
or
re.findall(r"((?:A|B|C)D)", "BDE")
I would like to know the reason behind the following behaviour:
>>> re.compile("(b)").split("abc")[1]
'b'
>>> re.compile("b").split("abc")[1]
'c'
I seems that when I add parentheses around the splitting pattern, re adds it into the split array. But why? Is it something consistent, or simply an isolated feature of regular expressions.
It's a feature of re.split, according to the documentation:
If capturing parentheses are used in pattern, then the text of all groups in the pattern are also returned as part of the resulting list.
In general, parenthesis denote capture groups and are used to extract certain parts of a string. Read more about capture groups.
In any regular expression, parentheses denote a capture group. Capture groups are typically used to extract values from the matched string (in conjunction with re.match or re.search). For details, refer to the official documentation (search for (...)).
re.split adds the matched groups in between the splitted values:
If capturing parentheses are used in pattern, then the text of all groups in the pattern are also returned as part of the resulting list.